Don't understand the answer of $\int_0^\pi \frac{\cos{n\theta}}{1 -2r\cos{\theta}+r^2} d\theta$

Question

I don't understand the answer provided in this post, what does "to use in the numerator $e^{inθ}$" mean? Also, how to handle the upper limit $\pi$ instead of $2\pi$? Many thanks in advance!

Answer 1

Since $e^{iz}=\cos z+i\sin z$, you have: $$\cos z=\frac{e^{iz}+e^{-iz}}{2}\quad\quad (*) $$ Using $(*)$, the numerator becomes: $$\cos n\theta=\frac{e^{in\theta}+e^{-in\theta}}{2} $$ Now, consider the denominator: $$ 1-2r\cos \theta+r^2=1-r(e^{i\theta}+e^{-i\theta})+r^2=e^{i\theta} e^{-i\theta}-re^{i\theta}-r e^{-i\theta}+r^2=\\ (e^{i\theta} -r)(e^{-i\theta}-r)=(e^{i\theta} -r)\left(\frac{1-re^{i\theta}}{e^{i\theta}}\right)=(1-re^{-i\theta})(1-re^{i\theta})$$